Did you find a different way to reach this value? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week!
Monday, August 31, 2015
Problem of the Week
We haven't done a geometry problem in a while, so here's one for this week's Problem of the Week.
Check out my written-up solution after the break to see if you got it right!
Did you find a different way to reach this value? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week!
Did you find a different way to reach this value? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week!
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I solved it via similar triangles, using the fact that the hypotenuse of the smaller right triangle is 1/cos(theta) from SOHCAHTOA. It gets the same answer, of course.
ReplyDeleteYour proof does not address what happens when theta is a right angle. Of course your formula gets the right result, but to be complete the proof should mention it.
i solved it with direction and orthogonal vectors, linked with the equation of l.
ReplyDeletesame answer !
If you look at the congruent right triangles each with the short side of length 1 which are generated by the radius of circle B along the x-axis and the radius along line m, you can write it as tan(1/2*(\pi - \theta)).
ReplyDelete